Question: Divide the following complex numbers. $ \dfrac{2-9i}{2+i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2-i}$ $ \dfrac{2-9i}{2+i} = \dfrac{2-9i}{2+i} \cdot \dfrac{{2-i}}{{2-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(2-9i) \cdot (2-i)} {(2+i) \cdot (2-i)} = \dfrac{(2-9i) \cdot (2-i)} {2^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(2-9i) \cdot (2-i)} {(2)^2 - (1i)^2} = $ $ \dfrac{(2-9i) \cdot (2-i)} {4 + 1} = $ $ \dfrac{(2-9i) \cdot (2-i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({2-9i}) \cdot ({2-i})} {5} = $ $ \dfrac{{2} \cdot {2} + {-9} \cdot {2 i} + {2} \cdot {-1 i} + {-9} \cdot {-1 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{4 - 18i - 2i + 9 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{4 - 18i - 2i - 9} {5} = \dfrac{-5 - 20i} {5} = -1-4i $